Timeline for Is there a proof of independence of AC from Z that is done in Z?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2022 at 9:35 | vote | accept | Zuhair Al-Johar | ||
| Sep 30, 2022 at 9:33 | comment | added | Zuhair Al-Johar | I think the relevant article is: Provident sets and rudimentary set forcing, by Mathias. Can be found at Researchgate: researchgate.net/publication/…. The answer to this question is affirmative! You pass to the provident closure of a model of Z, and work inside that model (Mathias). | |
| Sep 23, 2022 at 16:33 | comment | added | Noah Schweber | @ZuhairAl-Johar I could be wrong, but I don't believe that's the case (I haven't thought about it very hard though). | |
| Sep 23, 2022 at 15:23 | comment | added | Zuhair Al-Johar | hmmm.. I thought this can be done in Z+ ranks. | |
| Sep 23, 2022 at 14:55 | comment | added | Noah Schweber | @ZuhairAl-Johar Even making sense of $\nu[G]$ for a name $\nu$ needs replacement I think. | |
| Sep 23, 2022 at 14:10 | comment | added | Zuhair Al-Johar | I'm curious! Where do exactly Replacement is needed in forcing, I mean the names\values construction do not really needs the full power of Replacement, a theory like Z+ranks would be enough to do it, is it in the proofs about the forcing language which ensures the generic extension to be a model of the background theory?? Where exactly? | |
| Sep 22, 2022 at 19:15 | history | answered | Noah Schweber | CC BY-SA 4.0 |